(101)

Equations (100), (101) indicate that the amplitude of the time-averaged energy equals one half of the amplitude of the energy function given by Eq. (98) and is proportional to the frequency squared.

To obtain the power, substitute Eq. (97) into Eq. (83) and get

$P\left(x,t\right)=ce\left(x,t\right)=mc{\omega }^{2}{\stackrel{ˆ}{u}}^{2}{\mathrm{sin}}^{2}\left(\gamma x-\omega t\right)$ (102)

(102)

The power amplitude is

$\stackrel{ˆ}{P}=mc{\omega }^{2}{\stackrel{ˆ}{u}}^{2}$ (103)

(103)

To calculate the time-averaged power, recall Eq. (81), i.e.,

$P={\int }_{A}\stackrel{\to }{t}\cdot \stackrel{˙}{\stackrel{\to }{u}}dA=-\sigma \stackrel{˙}{u}A=$

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